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Difference between revisions of "1956 AHSME Problems/Problem 49"

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Also, <math>\angle BAT = 180^\circ - \angle BAP</math>, and <math>\angle ABR = 180^\circ - \angle ABP</math>. Hence,  
 
Also, <math>\angle BAT = 180^\circ - \angle BAP</math>, and <math>\angle ABR = 180^\circ - \angle ABP</math>. Hence,  
  
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<math>\angle AOB &= 180^\circ - \angle BAO - \angle ABO  
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<math>\angle AOB 180^\circ - \angle BAO - \angle ABO  
 
  180^\circ - \frac{\angle BAT}{2} - \frac{\angle ABR}{2}  
 
  180^\circ - \frac{\angle BAT}{2} - \frac{\angle ABR}{2}  
 
  180^\circ - \frac{180^\circ - \angle BAP}{2} - \frac{180^\circ - \angle ABP}{2}  
 
  180^\circ - \frac{180^\circ - \angle BAP}{2} - \frac{180^\circ - \angle ABP}{2}  

Revision as of 12:38, 1 August 2020

Solution 1

First, from triangle $ABO$, $\angle AOB = 180^\circ - \angle BAO - \angle ABO$. Note that $AO$ bisects $\angle BAT$ (to see this, draw radii from $O$ to $AB$ and $AT,$ creating two congruent right triangles), so $\angle BAO = \angle BAT/2$. Similarly, $\angle ABO = \angle ABR/2$.

Also, $\angle BAT = 180^\circ - \angle BAP$, and $\angle ABR = 180^\circ - \angle ABP$. Hence,

$\angle AOB 180^\circ - \angle BAO - \angle ABO 180^\circ - \frac{\angle BAT}{2} - \frac{\angle ABR}{2} 180^\circ - \frac{180^\circ - \angle BAP}{2} - \frac{180^\circ - \angle ABP}{2} \frac{\angle BAP + \angle ABP}{2}.$

Finally, from triangle $ABP$, $\angle BAP + \angle ABP = 180^\circ - \angle APB = 180^\circ - 40^\circ = 140^\circ$, so \[\angle AOB = \frac{\angle BAP + \angle ABP}{2} = \frac{140^\circ}{2} = \boxed{70^\circ}.\]

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